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Let $f:X\to Y$ be a local homeomorphism between Hausdorff spaces such that $f$ is surjective and $f^{-1}(y)$ is compact $\forall y\in Y$. If $|f^{-1}(y_1)|=|f^{-1}(y_2)|,\forall y_1,y_2\in Y$ then show that $f$ is a covering map.

Attempt: I know $f^{-1}(y)$ is finite as it has a finite cover $\cup_{i=1}^nU_i\supseteq f^{-1}(y)$ with each $U_i$ open such that $f_{U_i}$ is a homeomorphism. I considered $V_i\subseteq U_i$ with $V_i\cap V_j=\emptyset$ but the problem is the preimage of $\cap_{i=1}^nf(V_i)$ is not necessarily in $\cup_{i=1}^nU_i$ otherwise I would have found an evenly covered neighborhood of $y$. I need to show that $f$ is closed so I can use the set $(\cap_{i=1}^nf(V_i))\backslash (f(X\backslash \cup_{i=1}^nU_i))$ instead or consider another approach by using the constant preimage assumption which I do not see how it fits there...

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3 Answers 3

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It is much simpler than I thought it seems: the sets $V_i$ are the disjoint restrictions of the $U_i$ as defined in my question above. Suppose that $|f^{-1}(y)|=n,\forall y\in Y$. The problem is that taking the set $V=\cup_{i=1}^nf(V_i)$, I want to show that $f^{-1}(V)\subseteq \cup_{i=1}^nV_i$ so I can use the local 1-1 property and conclude that the preimages are exactly the open sets $f^{-1}(V)\cap U_i$. This is straghtforward however: if $y_0\in f^{-1}(V)$ then there exist $v_1\in V_1,...,v_n\in V_n: f(v_i)=y_0$ as we simply have $v_i=f_{V_i}^{-1}(y_0)$. But $|f^{-1}(y_0)|=n \implies f^{-1}(y_0)=\{v_1,...,v_n\}$ and thus $f^{-1}(y_0)\subseteq \cup_{i=1}^nV_i, \forall y_0\in Y$. The result now follows immediately.

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  • $\begingroup$ Yes, that seem correct. I guess the suggestion to prove the map closed led us all astray! $\endgroup$
    – Ruy
    Commented Dec 30, 2020 at 18:34
  • $\begingroup$ Indeed! We all miss some trivial results and take a longer route sometimes... $\endgroup$ Commented Dec 30, 2020 at 18:39
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Apparently the only remaining difficulty is to show that $f$ is closed, so let us concentrate in proving this.

So let $C$ be a closed subset of $X$ and pick $$ y\in \overline{f(C)}. $$ Write $$ f^{-1}(y)=\{x_1, x_2, \ldots , x_n\}, $$ and observe that $n$ is therefore the constant number of elements of each fiber of $f$.

For each $k=1,\ldots ,n$, choose open sets $U_k\subseteq X$ and $V_k\subseteq Y$, such that $x_k\in U_k$, $y\in V_k$, and $f$ is a homeomorphism from $U_k$ onto $V_k$. By using that $X$ is Hausdorff and shrinking the $U_k$ appropriately we may assume that the $U_k$ are pairwise disjoint.

Without loss of generality we may also assume that the $V_k$ are all the same since otherwise we may replace each $V_k$ by $$ V=\bigcap_{k=1}^n V_k, $$ while replacing each $U_k$ by $f^{-1}(V)\cap U_k$.

Since $y\in \overline{f(C)}$, we may find a net $\{c_i\}_{i\in I}\subseteq C$, such that $y=\lim_{i\in I}f(c_i)$. By discarding an initial segment of our net we may assume that $f(c_i)\in V$, for all $i$, so we may choose $d^k_i\in U_k$, such that $f(d^k_i)=f(c_i)$.

Due to the fact that the $U_k$ are pairwise disjoint we have that, for each $i$, the elements $$ d^1_i, d^2_i,\ldots , d^n_i $$ are all distinct, so there are precisely $n$ of them, and since they all lie in $f^{-1}\big (f(c_i)\big )$ we deduce from the hypothesis that $$ f^{-1}\big (f(c_i)\big ) =\{d^1_i, d^2_i,\ldots , d^n_i\}. $$

It follows that $$ c_i\in \{d^1_i, d^2_i,\ldots , d^n_i\}, $$ for every $i$, so the pigeonhole principle implies that there is a $k$ and a co-final subset $I_0\subseteq I$, such that $c_i=d^k_i$ for all $i\in I_0$. We then have that $$ x_k = \lim_{i\in I}d^k_i = \lim_{i\in I_0}d^k_i = \lim_{i\in I_0}c_i \in C, $$ so $$ y=f(x_k)\in f(C), $$ whence $f(C)$ is closed.

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  • $\begingroup$ I cannot find anything wrong so far, seems like a nice approach. That would imply that I do not need the same number of elements in each fiber but having all fibers $|f^{-1}(y)|\leq M$ for a specified number $M$ as pigeonhole principle would give me an $m\leq M$ and a cofinal subset $I_1\subseteq I$ with $c_i\in\{d_i^1,...,d_i^m\},\forall i\in I_1$ right? $\endgroup$ Commented Dec 30, 2020 at 17:15
  • $\begingroup$ I don't think a bound on the size of each inverse image would suffice. A counter example would be to take $X$ the disjoint union of $\mathbb R$ and $(0,\infty )$, $Y=\mathbb R$, and $f$ the identity map on each component of $X$. That the inverse images are all of the same size is also used in my argument leading up to $$ f^{-1}\big (f(c_i)\big ) =\{d^1_i, d^2_i,\ldots , d^n_i\}. $$ $\endgroup$
    – Ruy
    Commented Dec 30, 2020 at 17:38
  • $\begingroup$ I think one may get away with assuming that the function $$ y↦|f^{-1}(y)| $$ is upper semicontinuous. $\endgroup$
    – Ruy
    Commented Dec 30, 2020 at 17:42
  • $\begingroup$ Of course! Your argument is correct as far as I can tell, I upvoted it but won't accept because there is a much more direct approach I should have seen. Your answer helped me realize this $\endgroup$ Commented Dec 30, 2020 at 18:24
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You don't need to show that $f$ is closed.

Let $y \in Y$. Since the fiber $f^{-1}(y)$ is compact, then it is closed because $X$ is Hausdorff, also, since $f^{-1}(y)$ is finite, then it is discrete.

Say $f^{-1}(y) = \{x_1, x_2, \dots, x_n\}$, then by the Hausdorff property of $X$, we can find open sets $\{U_1, U_2, \dots, U_n\}$ such that $x_i \in U_i$ and $U_i \cap U_j = \phi$ for each $i \neq j$.

Since $f$ is a local homeomorphism, we can find for each $x_i$ an open neighborhood $V_i$ such that $f(V_i)$ is an open neighborhood of $y$ and that the restriction $f|_{V_i}$ is a homeomorphism.

This implies that $W_i = U_i\cap V_i$ is an open neighborhood of $x_i$ for all $i$ and that $f(W_i)$ is an open neighborhood of $y$ from the fact that $f|_{V_i}$ is a homeomorphism, and thus, maps open sets in the subspace $V_i$ (each $W_i$ is an open set in the subspace $V_i$) to open sets in the subspace $f(V_i)$ which are also open sets in $Y$. Also, each $f|_{W_i}$ is a homeomorphism.

Therefore, $B = \bigcap\limits_{i}^{n}f(W_{i})$ is an open neighborhood of $y$.

Moreover, let $A_i = f^{-1}|_{W_i}(B) = W_i \cap f^{-1}(B)$. Since $f|_{W_i}$ is a homeomorphism onto $f(W_i)$, then it is also a surjection and an injection, and from the fact that $A_i \subseteq W_i$, $f(A_i) = f|_{W_i}(A_i) = B$. So each $A_i$ is mapped homeomoprhically to $B$.

Also, $x_i \in A_i$ for all $i$, and thus, $f^{-1}(y) \subseteq \bigcup\limits_{i}^{n}A_i$. Hence, $\{y\} = f(f^{-1}(y)) \subseteq f(\bigcup\limits_{i}^{n}A_i) = \bigcup\limits_{i}^{n}f(A_i) = B$. In other words, $B$ is a neighborhood of $y$.

The problem now is that we are not sure whether $f^{-1}(B)$ is contained in $\bigcup\limits_{i}^{n}A_i$.

Let $x' \in f^{-1}(B)$, then $f(x') = y' \in B$. Therefore, $y' \in f(W_i)$ for all $i$. Hence, we can find $x'_i$ in each $W_i$ such that $f(x'_i) = y'$. Using the fact that $|f^{-1}(y_1)| = |f^{-1}(y_2)|$ for all $y_1 \neq y_2$ and that $|f^{-1}(y)| = n$, we conclude that $f^{-1}(y')=\{x'_1, x'_2, \dots, x'_n\} \subseteq \bigcup\limits_{i}^{n}A_i$. But $x'$ must be one of $\{x'_1, x'_2, \dots, x'_n\}$, and thus, $x' \in \bigcup\limits_{i}^{n}A_i$ which completes the proof.

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